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The first and the last terms of an AP are 5 and 45 respectively. If the sum of all its terms is 400, Find its common difference.
Given: An A.P. , first term and last term 5 and 45 respectively. sum of all its terms$=400$.
To do: To find its common difference.
Solution:
Let $a$ be the first term and $d$ be the common difference.
Let us say the given A.P. has $n$ terms.
here as given,
first term $a=5$
last term $l=45$
sum of the A.P. $S_{n}=400$
$n=?$
As the given condition,
last term $l=a_{n}=a+(n-1)d$
$\Rightarrow 45=5+(n-1)d$
$\Rightarrow ( n-1) d=40\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ .................( 1)$
And sum of all its terms $S_{n} =\frac{n}{2}[ 2a+( n-1) d]$
$\Rightarrow \frac{n}{2}[ 2\times 5+40] =400 \ \ \ \ \ ( \because ( n-1) d=40\ from\ ( 1)$
$\Rightarrow 50n=800$
$\Rightarrow n=\frac{800}{50} =16$
$n=16$, on subtituting the value of $n$ into $( 1)$ ,
$( 16-1) d=40$
$\Rightarrow 15d=40$
$\Rightarrow d=\frac{40}{15}$
$\Rightarrow d=\frac{8}{3}$
Therefore the common difference of the given A.P. is $\frac{8}{3} $.
Updated on: 10-Oct-2022
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